Question

The magnetic field associated with a light wave is given, at the origin, by $B=B_{0} \, \left[sin \left(3.14 \times \left(10\right)^{7}\right) c t + sin ⁡ \left(6.28 \times \left(10\right)^{7}\right) c t\right].$ If this light falls on a silver plate having a work function of $4.7eV,$ what will be the maximum kinetic energy of the photoelectrons?

$\left(\right.c=3\times \left(10\right)^{8} \, m \, s^{- 1},h=6.6\times \left(10\right)^{- 34} \, J \, s\left.\right)$

• A. $6.82 \, eV$
• B. $7.72 \, eV$
• C. $12.5 \, eV$
• D. $8.52eV$

Answer 1

Answer: (B)

$B=B_{0}\left[sin \left(3.14 \times \left(10\right)^{7} c t\right) + sin ⁡ \left(6.28 \times \left(10\right)^{7} c t\right)\right]$
Maximum $KE$ will associate with the maximum frequency
$f_{max \, }=\frac{6.28 \times 10^{7} c}{2 \pi }=10^{7}\times 3\times 10^{8}Hz=3\times 10^{15} \, Hz$
$KE_{m a x}=hf_{m a x}-\phi$
$=\left[\frac{6 .63 \times 10^{- 34} \times 3 \times 10^{15}}{1 .6 \times 10^{- 19}} - 4 .7 \, eV\right]$
$=7.72 \, eV$